Lemma. Perfect counting duality [fdip-001F]
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  • October 20, 2026

Let \(R\) be a commutative unital ring, \(V\) a finite set and \(H=\mathcal F_R(V)\) its field module. The counting bilinear form is \[b(f,g)=\sum _{v\in V}f(v)g(v).\] It is symmetric. Write \(H^\vee =\operatorname {Hom}_R(H,R)\) for the dual module, using ordinary \(R\)-linear maps: both coefficient rings are \(R\) and \(\sigma =\operatorname {id}_R\), as in Remark [ca-001G]. The perfect counting duality is the pair of inverse \(R\)-linear maps \[\begin {aligned} \Phi :H&\longrightarrow H^\vee ,&\Phi (g)(f)&=b(f,g),\\ \Psi :H^\vee &\longrightarrow H,&(\Psi \ell )(v)&=\ell (\delta _v). \end {aligned}\] Here \(\delta _v(u)\) is \(1\) when \(u=v\), and \(0\) otherwise. Perfection means precisely that \(\Phi \) is an isomorphism.

Proof. 

  • October 20, 2026

Commutativity and distributivity in \(R\) give symmetry and bilinearity of \(b\); they also give linearity of \(\Phi \) and \(\Psi \). For \(f\in H\), evaluation at every point proves \(f=\sum _v f(v)\delta _v\). Consequently \[(\Phi \Psi \ell )(f) =\sum _v f(v)\ell (\delta _v) =\ell \left (\sum _v f(v)\delta _v\right )=\ell (f).\] Conversely, \((\Psi \Phi g)(v)=b(\delta _v,g)=g(v)\). Thus both composites are identities.

This proof includes the empty set and the zero ring. It uses no order, positivity or division. A perfect form can have nonzero vectors of zero self-pairing: over \(\mathbb F_2\), the counting form on two points has \(b((1,1),(1,1))=0\). Conversely, separation of vectors alone does not imply perfection over a ring: the form \(b(x,y)=2xy\) on \(\mathbb Z\) cannot represent the functional \(x\mapsto x\). The Hom-module convention is [stacks2017stacks, Section 10.10], Tag 0581; the explicit inverse above supplies the finite counting assertion.